Apply the relationships for uniform circular motion, including centripetal acceleration a = v^2/r, centripetal force F = m v^2 / r, period T = 2 pi r / v, and the geometry of banked curves and conical pendulums

What this dot point is asking

QCAA wants you to apply the kinematic and dynamic relationships for an object moving in a circle at constant speed, identify the real force (or component of a real force) supplying the centripetal force in a given context, and solve geometry problems for banked curves and conical pendulums. The dot point appears in IA1 short response, often as a banked-curve stimulus, and is a popular IA2 design (conical pendulum or rotating-mass setup).

The answer

Uniform circular motion in one paragraph

An object moving at constant speed $v$ around a circle of radius $r$ has changing velocity (because direction is changing). The acceleration is directed toward the centre of the circle (centripetal):

By Newton's second law, this requires a net inward force:

Centripetal force is not a separate kind of force. It is the role played by whichever real force (or component of one) points toward the centre. In different contexts:

Period and frequency

For one complete revolution the object travels a distance $2 \pi r$ at speed $v$, so the period is:

and the frequency is $f = 1/T$. Angular speed is $\omega = 2 \pi f = v/r$, so $a_c = \omega^2 r$ is an equivalent form.

Banked curves

On a curve banked at angle $\theta$ to the horizontal, the normal force is perpendicular to the road surface. Resolving:

• Vertical: $N \cos\theta = m g$ (no vertical acceleration).
• Horizontal: $N \sin\theta = m v^2 / r$ (centripetal).

Dividing eliminates $N$ and $m$:

This is the design speed of the banked curve. At that speed, no friction is required to keep the vehicle on the curve. Above the design speed, friction must act down the slope (toward the inside of the curve); below it, friction must act up the slope.

Try it: Banking angle calculator. Enter speed and radius to get the design angle and the friction required at any other speed.

Conical pendulum

A mass swung in a horizontal circle on a string at angle $\theta$ from the vertical has tension components:

• Vertical: $T \cos\theta = m g$.
• Horizontal: $T \sin\theta = m v^2 / r$.

The relationship between angle and speed is identical to the banked curve: $\tan\theta = v^2 / (r g)$. The radius of the horizontal circle is $r = L \sin\theta$ where $L$ is the string length.

The period of a conical pendulum:

A classic IA2 measures period as a function of cone angle (or string length) and linearises against $\cos\theta$ (or $L$) to extract $g$.

Worked example: car on a flat bend

A 900 kg car rounds a flat bend of radius 30 m at 12 m/s. The required centripetal force is $F_c = 900 \times 12^2 / 30 = 4320$ N, supplied entirely by static friction.

Try it: Centripetal force calculator. Enter mass, speed and radius for $F_c$ and $a_c$.

How this appears in IA1 and IA2

IA1 data test. Expect a banked-curve diagram with the bank angle and a speed, asked to determine whether the vehicle stays on the curve, to find the design speed, or to extract the friction coefficient required at a non-design speed. Alternatively, a conical pendulum stimulus with multiple period measurements at different angles.

IA2 student experiment. Common designs: a conical pendulum with mass, period and angle measured to test $T = 2 \pi \sqrt{L \cos\theta / g}$; a rotating-mass setup on a horizontal turntable with friction varied; a ball-in-a-cone or marble-in-a-bowl rolling at different heights. Strong reports linearise the relationship before fitting (e.g. $T^2$ vs $\cos\theta$ slopes give $4\pi^2 L / g$) and propagate uncertainty cleanly.

Examples in context

Example 1. A Bremer River bridge approach has a curved deck of radius $r = 80 \text{ m}$ banked at $\theta = 6^\circ$. The frictionless design speed is $v = \sqrt{rg\tan\theta} = \sqrt{80 \times 9.8 \times 0.105} = 9.07 \text{ m s}^{-1}$ ($32.7 \text{ km h}^{-1}$). Cars at the posted $60 \text{ km h}^{-1}$ exceed this and rely on tyre friction up the bank for the extra centripetal force. QCAA Unit 3 EA Paper 2 banked-curve calculations follow this template exactly.

Example 2. A Cairns hospital centrifuge rotor at $4000 \text{ rpm}$ ($\omega = 419 \text{ rad s}^{-1}$) spins samples at $r = 0.10 \text{ m}$. Centripetal acceleration is $a = \omega^2 r = 1.76 \times 10^4 \text{ m s}^{-2}$, equivalent to $1790g$. The supporting force from the rotor cup wall is $F = ma$; for $m = 0.005 \text{ kg}$ samples this is $88 \text{ N}$. The dot-point relation $a = v^2/r$ also applies via $v = \omega r$.

Try this

Q1. State the centripetal acceleration and force formulas. [2 marks]

• Cue. $a = v^2/r$; $F = mv^2/r$, directed toward the centre.

Q2. A $1200 \text{ kg}$ car rounds a flat horizontal curve of radius $50 \text{ m}$ at $15 \text{ m s}^{-1}$. Calculate the required centripetal force and identify the source. [3 marks]

• Cue. $F = mv^2/r = 1200 \times 225/50 = 5400 \text{ N}$; supplied by tyre friction.

Q3. A Bremer River bridge banked approach: $r = 80 \text{ m}$, $\theta = 6^\circ$. (a) Calculate the frictionless design speed. (b) Determine the friction force on a $1500 \text{ kg}$ car at $20 \text{ m s}^{-1}$ ($72 \text{ km h}^{-1}$). (c) Discuss one safety implication during wet-weather. [3+3+2 marks; ISMG: Analysis and interpretation, Evaluation]

• Cue. (a) $9.07 \text{ m s}^{-1}$; (b) friction up-bank component supplies remainder of $F_c = mv^2/r = 7500 \text{ N}$ above banking; (c) reduced $\mu$ lowers safe max speed.